Question

Considere o polinômio de grau 3

P(x) = ax^3 + bx^2 + cx + d

(a diferente de zero).

Reescreva-o como a soma de dois cubos de polinômios de grau 1, mais uma constante.

Answer 1

Olá Lukyo.



Produtos notáveis usados:

$\star~\boxed{\boxed{\mathsf{(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3}}}\\\\\\\star~\boxed{\boxed{\mathsf{(a\pm b)^2=a^2\mp 2ab+b^2}}}\\\\\\\star~\boxed{\boxed{\mathsf{a^2\pm b^2=(a\pm b)\cdot(a\mp b)}}}$

Pelo enunciado queremos reescrever P(x) da seguinte forma.

$\mathsf{P(x)=ax^3+bx^2+cx+d=(a'x+g)^3+(a''x+l)^3+C}$

Mas antes vamos dividir somente o polinômio de terceiro grau por a/2, onde obteremos o polinômio G(x), e igualar a soma de dois cubos mais uma constante.

$\mathsf{G(x)=2x^3+\dfrac{2bx^2}{a}+\dfrac{2cx}{a}+\dfrac{2d}{a}=(a'x+g)^3+(a''x+l)^3+C}$

$\mathsf{2x^3+\dfrac{2bx^2}{a}+\dfrac{2cx}{a}+\dfrac{2d}{a}=\Big(a'x+g\Big)^3+\Big(a''x+l\Big)^3+C}$


Desenvolvendo o lado direito, temos:

$\mathsf{(a'x+g)^3+(a''x+l)^3+C}\\\\\\\mathsf{(a'x)^3+3(a'x)^2g+3(a'x)g^2+g^3+(a''x)^3+3(a''x)^2l+3(a''x)l^2+l^3+C}\\\\\\\mathsf{x^3(a'^3+a''^3)+x^2(3a'^2g+3a''^2l)+x(3a'g^2+3a''l^2)+g^3+l^3+C}$

Comparando termo a termo com a equação polinomial de terceiro grau.

$\mathsf{\diagup\!\!\!\!x^3(a'^3+a''^3)=2\diagup\!\!\!\!x^3}\\\\\mathsf{a'^3+a''^3=2}$

Precisamos agora escrever 2 como a soma de dois cubos. Uma solução trivial seria a' e a'' igual a 1.

$\mathsf{2=1^3+1^3}$

Comparando os demais termos:

$\mathsf{\diagup\!\!\!\!x^2(3a'^2g+3a'^2l)=\dfrac{2b\diagup\!\!\!\!x^2}{a}}\\\\\mathsf{3\cdot(1^2g+1^2l)=\dfrac{2b}{a}}\\\\\mathsf{g+l=\dfrac{2b}{3a}}~~(i)\\\\\\\\\mathsf{\diagup\!\!\!\!x(3a'g^2+3a''l^2)=\dfrac{2c\diagup\!\!\!\!x}{a}}\\\\\mathsf{3\cdot(g^2+l^2)=\dfrac{2c}{a}}\\\\\mathsf{g^2+l^2=\dfrac{2c}{3a}~~}(ii)$

Vamos isolar agora uma das contantes g ou l em (i) e substituir em (ii).

$\mathsf{g+l=\dfrac{2b}{3a}}\\\\\mathsf{(g)^2=\Big(\dfrac{2b}{3a}-l\Big)^2}\\\\\mathsf{g^2=\Big(\dfrac{4b^2}{9a^2}-\dfrac{4bl}{3a}+l^2\Big)}$

Substituindo g² em (ii).

$\mathsf{g^2+l^2=\dfrac{2c}{3a}~~(ii)}\\\\\\\mathsf{\Big(\dfrac{4b^2}{9a^2}-\dfrac{4b}{3a}+l^2\Big)+l^2=\dfrac{2c}{3a}}\\\\\\\mathsf{2l^2-\dfrac{4bl}{3a}=\dfrac{2c}{3a}-\dfrac{4b^2}{9a^2}~~\cdot(2)}\\\\\\\mathsf{4l^2-\dfrac{8bl}{3a}=\dfrac{4c}{3a}-\dfrac{8b^2}{9a}~~+\Big(\dfrac{2b}{3a}\Big)^2}\\\\\\\mathsf{(2l)^2-\dfrac{2\cdot2b\cdot2l}{3a}+\Big(\dfrac{2b}{3a}\Big)=\dfrac{12ac}{9a^2}-\dfrac{8b^2}{9a}+\dfrac{4b^2}{9a^2}}\\\\\\\mathsf{(2l-\dfrac{2b}{3a})^2=\dfrac{12ac-4b^2}{9a^2}}\\\\\\\mathsf{2l-\dfrac{2b}{3a}=\pm\sqrt{\dfrac{4\cdot(3ac-b^2)}{9a^2}}}$

$\mathsf{2l-\dfrac{2b}{3a}=\pm\sqrt{\dfrac{4\cdot(3ac-b^2)}{9a^2}}}\\\\\\\mathsf{2l=\pm\dfrac{2\sqrt{3ac-b^2}}{3a}+\dfrac{2b}{3a}~~\cdot\Big(\dfrac{1}{2}\Big)}\\\\\\\mathsf{l=\pm\dfrac{\sqrt{3ac-b^2}+b}{3a}}$

Se tivéssemos isolado o l em (i) ao invés de g, a solução seria a mesma. Portanto serão simétricos, onde um será positivo e o outro negativo.

Por último, a constante C.

$\mathsf{C+g^3+l^3=\dfrac{2d}{a}}\\\\\\\mathsf{C+\Big(\dfrac{\sqrt{3ac-b^2}+b}{3a}\Big)^3+\Big(\dfrac{-\sqrt{3ac-b^2}+b}{3a}\Big)^3=\dfrac{2d}{a}}\\\\\\\mathsf{C+\Big(\dfrac{2b}{3a}\Big)\!\cdot\!\Big[\Big(\dfrac{\sqrt{3ac\!-\!b^2}+b}{3a}\Big)^2\!\!-\Big[\dfrac{b^2\!-\!(3ac\!-b^2)}{9a^2}\Big]\!\!+\!\Big(\dfrac{-\sqrt{3ac-b^2}+b}{3a}\Big)^2\Big]}\\\\\\\mathsf{C+\Big(\dfrac{2b}{3a}\Big)\cdot\Big[\dfrac{-2b^2+9ac}{9a^2}\Big]=\dfrac{2d}{a}}\\\\\\\mathsf{C=\dfrac{2d}{a}-\Big(\dfrac{-4b^3+18abc}{27a^3}\Big)}$

$\mathsf{C=\dfrac{54da^2+4b^3-18abc}{27a^3}}$

Portanto G(x) ficaria:


$\mathsf{G(x)=\Big(x+\dfrac{\pm\sqrt{3ac-b^2}+b}{3a}\Big)^3+\Big(x+\dfrac{\mp\sqrt{3ac-b^2}+b}{3a}\Big)^3+C}$


Mas como queremos P(x) escrito como a soma de dois cubos mais uma constante, precisamos multiplicar G(x) por a/2.


$\boxed{\mathsf{P(x)=\dfrac{a}{2}\cdot\Big[\Big(x+\dfrac{\pm\sqrt{3ac-b^2}+b}{3a}\Big)^3+\Big(x+\dfrac{\mp\sqrt{3ac-b^2}+b}{3a}\Big)^3+C}\Big]}}$



Dúvidas? comente.